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Let n be a positive integer and let $H=\{z\in\mathbb{C} | \space \space z^n = 1 \} $

i)Prove that the sum of all elements of H is $0$

ii) Prove that the product of all the elements of H is 1 if n is odd and -1 if n is even.

I)Picking some fixed n and drawing this out its very obvious when n is even and still rather obvious when n is odd but how do i write it out formally which holds for any fixed n?

II) when this is odd we can pair the second and last the third and second last etc together each product will =1 as they are inverse of each other and the first term is $e^{0}=1$ for the even case i have no idea.

$H=<e^{\frac{i2\pi k}{n}}>$ for some fixed postive integer n where k is in $\mathbb{Z}$

Faust
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  • Geometric intuition will help here. When you draw it out, why does the claim become obvious? – Chris Culter Apr 02 '17 at 04:45
  • when its even there is a reflection about the y axis vector for ever vector on the right hand side clearly there sum = 0 – Faust Apr 02 '17 at 04:46
  • Okay, that's a start! Now, can you write out that argument in algebra? – Chris Culter Apr 02 '17 at 04:55
  • only for the even case my problem is with the odd case i need to sum the 2 specific elements that are not reflected like in the even case and show there sum going left equal the first element going right and there up down cancel. but in math not gibberish ^^ – Faust Apr 02 '17 at 04:57
  • Well, in the odd case, there is no pairing of elements that sum to zero. Is there anything else you can try? (Geometrically, I mean.) Let's say specifically for $n=5$. What makes it obvious visually in that case? – Chris Culter Apr 02 '17 at 05:06
  • I dunno it just looks like it O.o the up down pairs of the 4 not pointing in just the x only direction appear to cancel. intuition from the even case would like the remainder pointing left to cancel with the amount pointing right... if you draw it with vector arrows adding them with correct distances they do cancel to 0 could i somehow sum the angles together? – Faust Apr 02 '17 at 05:13
  • Great! The up-down pairs are called complex conjugates. For each $z\in H$, we also have its complex conjugate $\bar z\in H$. So if $s$ is the sum, then $\bar s=s$, which means the imaginary part of $s$ is $0$. Now we need to deal with the real part. Is there any other symmetry of $H$ that you can use? – Chris Culter Apr 02 '17 at 05:17
  • Seems like a much better way of approaching it but i don't see anything else that one could generalize to an n case. thanks anyway though. – Faust Apr 02 '17 at 05:25
  • Hint: Rotation... – Chris Culter Apr 02 '17 at 05:40
  • My picture must be wrong for that to make any sense so i should goto sleep will have anthor look at it again tommorrow. thx. – Faust Apr 02 '17 at 05:47
  • Okay, see this answer: http://math.stackexchange.com/a/1806750/87023 but any of the other answers will also do. Note that your original claim isn't true as stated; you need $n>1$. – Chris Culter Apr 02 '17 at 05:47

1 Answers1

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1) Geometry is the best approach for this problem, unless you know some field theory or complex analysis. The points are equally arranged around the unit circle. The mean value of the points is equal to their sum, divided by $n$. Since the points are evenly spaced around the unit circle, their mean value is $0$. This is because the mean value of the numbers is equal to the mean position of the points, which in this case is the center of the circle. Therefore their sum is also $0$.

2) Your argument is almost done! For the even case, the same thing happens except $-1$ is left out. Thus for the even case, the product isn't $1$, but rather is $-1$ as desired.

Also, see Vieta's Formulas which answer both questions.

Stella Biderman
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  • hmm which element is the $-1$ case when n is even? i think ill want to use the same argument on n-2 terms then pair the identity with the term that always becomes -1. is it the middle term after removing the identity ? oh i think i remember this argument whatever the element is it should be its own inverse! – Faust Apr 02 '17 at 04:53
  • $e^{\pi i}=-1$ is a solution to the equation $z^{2k}=1$ for every positive integer $k$. – Stella Biderman Apr 02 '17 at 04:55
  • @Faust7 Yes, that pairing method works. Marching everything with its inverse leaves out only the numbers that are their own inverses. You'll have to prove that the only ones that are their own inverses are $\pm 1$ to finish the proof. – Stella Biderman Apr 02 '17 at 04:59
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    ah if we remove the first element and arrange them in order it is the middle one. i read your Vietas formula link but it went a little over my head. i think once im done with this ill go over it again. i know an argument to show that there is 2 things that are there own inverses cause were in a cyclic group but im not sure how to show its those two elements. can i simply multiply them together and show that there there own inverses? – Faust Apr 02 '17 at 04:59
  • @Faust7 The general form is a little complicated, but the key bit is that for a polynomial with leading coefficient equal to $1$, the constant coefficient is the product of the roots multiplied by $(-1)^d$ where $d$ is the degree of the polynomial, and the second to largest degree coefficient is the negative of the sum of the roots (the largest degree coefficient is $1$, by assumption). Applying these two statements to $z^n-1$ solves both of your problems instantly. – Stella Biderman Apr 02 '17 at 05:03